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The absolute value (or modulus) \(|x|\) of a real number x is x's numerical value without regard to its sign.

For example, \(|3| = 3\); \(|-12| = 12\); \(|-1.3|=1.3\)

Graph:

Important properties:

\(|x|\geq0\)

\(|0|=0\)

\(|-x|=|x|\)

\(|x|+|y|\geq|x+y|\)

\(|x|\geq0\)

How to approach equations with moduli

It's not easy to manipulate with moduli in equations. There are two basic approaches that will help you out. Both of them are based on two ways of representing modulus as an algebraic expression.

1) \(|x| = \sqrt{x^2}\). This approach might be helpful if an equation has × and /.

2) |x| equals x if x>=0 or -x if x<0. It looks a bit complicated but it's very powerful in dealing with moduli and the most popular approach too (see below).

3-steps approach:

General approach to solving equalities and inequalities with absolute value:

1. Open modulus and set conditions.To solve/open a modulus, you need to consider 2 situations to find all roots:

Positive (or rather non-negative)

Negative

For example, \(|x-1|=4\)a) Positive: if \((x-1)\geq0\), we can rewrite the equation as: \(x-1=4\)b) Negative: if \((x-1)<0\), we can rewrite the equation as: \(-(x-1)=4\)We can also think about conditions like graphics. \(x=1\) is a key point in which the expression under modulus equals zero. All points right are the first condition \((x>1)\) and all points left are second condition \((x<1)\).

2. Solve new equations:a) \(x-1=4\) --> x=5b) \(-x+1=4\) --> x=-3

3. Check conditions for each solution:a) \(x=5\) has to satisfy initial condition \(x-1>=0\). \(5-1=4>0\). It satisfies. Otherwise, we would have to reject x=5.b) \(x=-3\) has to satisfy initial condition \(x-1<0\). \(-3-1=-4<0\). It satisfies. Otherwise, we would have to reject x=-3.

3-steps approach for complex problems

Let’s consider following examples,

Example #1Q.: \(|x+3| - |4-x| = |8+x|\). How many solutions does the equation have?Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:

a) \(x < -8\). \(-(x+3) - (4-x) = -(8+x)\) --> \(x = -1\). We reject the solution because our condition is not satisfied (-1 is not less than -8)

(Optional) The following illustration may help you understand how to open modulus at different conditions.Answer: \(-\sqrt{5}\), \(-\sqrt{3}\), \(\sqrt{3}\), \(\sqrt{5}\)

Tip & Tricks

The 3-steps method works in almost all cases. At the same time, often there are shortcuts and tricks that allow you to solve absolute value problems in 10-20 sec.

I. Thinking of inequality with modulus as a segment at the number line.

For example,Problem: 1<x<9. What inequality represents this condition?A. |x|<3B. |x+5|<4C. |x-1|<9D. |-5+x|<4E. |3+x|<5Solution: 10sec. Traditional 3-steps method is too time-consume technique. First of all we find length (9-1)=8 and center (1+8/2=5) of the segment represented by 1<x<9. Now, let’s look at our options. Only B and D has 8/2=4 on the right side and D had left site 0 at x=5. Therefore, answer is D.

II. Converting inequalities with modulus into a range expression.In many cases, especially in DS problems, it helps avoid silly mistakes.

For example,|x|<5 is equal to x e (-5,5).|x+3|>3 is equal to x e (-inf,-6)&(0,+inf)

III. Thinking about absolute values as the distance between points at the number line.

For example, Problem: A<X<Y<B. Is |A-X| <|X-B|?1) |Y-A|<|B-Y|Solution:We can think of absolute values here as the distance between points. Statement 1 means than the distance between Y and A is less than that between Y and B. Because X is between A and Y, |X-A| < |Y-A| and at the same time the distance between X and B will be larger than that between Y and B (|B-Y|<|B-X|). Therefore, statement 1 is sufficient.

Pitfalls

The most typical pitfall is ignoring the third step in opening modulus - always check whether your solution satisfies conditions.

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26 Nov 2017, 07:23

I am Able to solve most of the absolute value problems using the 3 steps approach. But every time I am taking more than 2 minutes to solve them. Is it possible to apply number line method to solve all the modulus problems? Can anyone please explain with some examples how to apply number line method.

Ok, Now after literally banging my head for 3 hrs and reading you blog articles back and forth, I get it that to make an EQ in (X-K) format we manipulate it by taking -tive sign outbut I guess in this example its this concept that we need to understand

|x|= x when x is >= 0, |x|= -x when x < 0

ok, so based on this understanding I will take a fresh shot, please let me know what's wrong

Quote:

a) \(x < -8\). \(-(x+3) - (4-x) = -(8+x)\) --> \(x = -1\). We reject the solution because our condition is not satisfied (-1 is not less than -8)

In this test case, since we will always have |x+3| negative we put a -tive sign outside because modulus will turn it into non negative, so to do that we take multiply it by (-1), is this understanding correct?and since we are ok with -(4-x), because we will again get |4-x| positive with a negative x, the -tive sign outside the bracket will make sure its always -tive when out of the Modulus. However to be frank, a little confusion here is, as you mentioned in the blog, why don't we try to convert it into (x-k) format?in RHS we have -(8-x) because again we want |8-x| to turn out a negative number so we put -(8-x) to make it always negative, let me know if I got it correctly.

I still dont get it, if we test x against both -tive and positive scenario, why is that we just have 1 equation? in my view we should split it in 2 eq. to test against both negative and positive value.

Quote:

d) \(x \geq 4\). \((x+3) + (4-x) = (8+x)\) --> \(x = -1\). We reject the solution because our condition is not satisfied (-1 is not more than 4)

(x+3) since we don't have a negative value of X this bracket will always be positive, we don't need a -tive sign outside, is this the reason?(4-x) again since a >4 will always make it positive we don't need a -tive sign outside the bracket, is this the reason?(8+x) again same reason as above for this?

First of all, if you do get a question with multiple mods and if you want to be prepared for it, using algebra will be far more time consuming than the approaches discussed in my blog. But nevertheless, you should understand it properly.

When you have an equation with x in it, you solve by taking x to one side and everything else to the other. What happens when you have mods in it?Say |x| = 4, you still haven't got the value of x. You have the value of |x| only. So you need to remove the mod. Now there are rules to remove the mod.

|x|= x (mod removed) when x is >= 0, |x|= -x (mod removed) when x < 0

So |x| = 4 to remove the mod, I need to know whether x is positive or negative.If x >= 0, |x| = x so |x| = 4 = xWe get that x is 4

If x < 0, |x| = -x so |x| = 4 = -x hence x = -4

So if we are looking for a positive value, then it is 4 and if we are looking for a negative value, it is -4.

Similarly, when you have |x+4| + |x - 3| = 10 (just an example), you need to remove the mods to solve for x. But to remove mods (which are around the entire factors x-4 and x-3 and not just around x), you need to know whether (x+ 4) and (x - 3) (the thing inside the mod) are positive/negative.

you say ' Similarly, when you have |x+4| + |x - 3| = 10 (just an example), you need to remove the mods to solve for x. But to remove mods (which are around the entire factors x-4 and x-3 and not just around x), you need to know whether (x+ 4) and (x - 3) (the thing inside the mod) are positive/negative.

Can you square |x+4| + |x - 3| = 10?Yes you can. It is an equation. You can always square an equation. Even if it were an inequality such as |x+4| + |x - 3| > 10, you can still square it since both sides are definitely non negative. But what will we achieve by squaring it? It doesn't remove the absolute value sign.

\(|x+4| + |x - 3| = 10\)

\(|x + 4|^2 + |x - 3|^2 + 2*|x+4|*|x - 3| = 10^2\)

The term 2*|x+4|*|x - 3| will continue to have absolute value sign. How will you take care of it?
_________________

Can you square |x+4| + |x - 3| = 10?Yes you can. It is an equation. You can always square an equation. Even if it were an inequality such as |x+4| + |x - 3| > 10, you can still square it since both sides are definitely non negative. But what will we achieve by squaring it? It doesn't remove the absolute value sign.

\(|x+4| + |x - 3| = 10\)

\(|x + 4|^2 + |x - 3|^2 + 2*|x+4|*|x - 3| = 10^2\)

The term 2*|x+4|*|x - 3| will continue to have absolute value sign. How will you take care of it?

i extracted his solution here (in wine red with question stem and answer choices ), so you dont overload your laptop with many internet browser tabs opened otherwise your laptop risks to be frozen like mine

so see below this magician squares modulus so my question when can i apply squaring to modulus ? in which cases can I apply and which cases it wont work ?

How many values of x satisfy the equation ||x-7|-9|=11?

A. 1 B. 2 C. 3 D. 4 E. 5

If |x-7| = a, the equation will become |a - 9| = 11 ( how did he break this equation into two btw ?

Squaring on both sides, \(a^2\)−18a+81=121\(a^2\)−18a+81=121

\(a^2\)−18a−40=0\(a^2\)−18a−40=0

Solving for a, a = 20 or -2

If |x-7| = 20x could take the value 27 or -13However, it is not possible to get a value of x where |x-7| = -2.

Therefore, there are 2 values of x which satisfy the equation(Option B)

Can you square |x+4| + |x - 3| = 10?Yes you can. It is an equation. You can always square an equation. Even if it were an inequality such as |x+4| + |x - 3| > 10, you can still square it since both sides are definitely non negative. But what will we achieve by squaring it? It doesn't remove the absolute value sign.

\(|x+4| + |x - 3| = 10\)

\(|x + 4|^2 + |x - 3|^2 + 2*|x+4|*|x - 3| = 10^2\)

The term 2*|x+4|*|x - 3| will continue to have absolute value sign. How will you take care of it?

i extracted his solution here (in wine red with question stem and answer choices ), so you dont overload your laptop with many internet browser tabs opened otherwise your laptop risks to be frozen like mine

so see below this magician squares modulus so my question when can i apply squaring to modulus ? in which cases can I apply and which cases it wont work ?

How many values of x satisfy the equation ||x-7|-9|=11?

A. 1 B. 2 C. 3 D. 4 E. 5

If |x-7| = a, the equation will become |a - 9| = 11 ( how did he break this equation into two btw ?

Squaring on both sides, \(a^2\)−18a+81=121\(a^2\)−18a+81=121

\(a^2\)−18a−40=0\(a^2\)−18a−40=0

Solving for a, a = 20 or -2

If |x-7| = 20x could take the value 27 or -13However, it is not possible to get a value of x where |x-7| = -2.

Therefore, there are 2 values of x which satisfy the equation(Option B)

Note here that in this case squaring will take care of the absolute value sign since the whole expression on the left is within the sign.

So the same case is here. When you have \(a = |x + 4|\) and \(b = |x - 3|\), the 2ab term retains the absolute signs and hence you cannot solve it. To solve an equation, you need to get rid of all absolute value signs.

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